1/9
Limits at a point (1.8)
Limit of a function at a point (informal).
How do the values of f behave when x approaches some (finite) number c. What
is the limit of f ( x ) = 2x !1 as x approaches 3? Here the function is continuous at c
( c = 3) , and lim ( 2x " 1) = f ( 3) = 5 .
x! 3
sin x
as x approaches 0? Here the function is not
x
continuous at c ( c = 0 ) , but does the limit still exist?
x
-1
-0.1
-0.01
0
0.01
0.1
1
0.84147 0.99833 0.99998
-0.99998 0.99833 0.84147
sin x
x
sin x
From the table, it appears that lim
= 1 . What does this imply about sine
x!0
x
values for very small values of x? E.g., estimate sin (.001) .
Limit of a function (formal).
Assume that f ( x ) is defined for all x near c (i.e., in some open interval containing
c), but not necessarily at c itself. We say that the limit of f ( x ) as x approaches c is equal
to L if f ( x ) ! L becomes arbitrarily small when x is any number sufficiently close (but
not equal) to c. In this case, we write lim f ( x ) = L . We also say that f ( x ) approaches
What is the limit of g( x ) =
x!c
or converges to L ( f ( x ) ! L ) as x approaches c. Note that f ( c ) does not need to be
defined for the lim f ( x ) to exist.
x!c
EXAMPLES. Find the following limits, if they exist. DRAW graphs.
1
1
lim (break)!!!!!!!!!!!!!!!!!!!!!!!!!! lim 2 (break)
x!0 x
x!0 x
sin ( x )
x2
lim
!!( hole)!!!!!!!!!!!!!!!!!!!!lim
(hole)
x!0
x!0 x
x
x
lim
(jump)
x!0 x
" 1%
lim sin $ ' !!( oscillates )
x!0
# x&
1
(continuous)
x!1 x
When do limits exist? From the above examples, limits exist when the function:
1. Is continuous at the point the limit is being evaluated
2. Has a "remvovable" discontinuity (i.e. a hole) at the point the limit is
being evaluated
Limits do not exist when the function:
1. Has a "non-removable" discontinuity (i.e. a break or a jump) at the point
the limit is being evaluated
2. Oscillates between two different values
lim
One-sided limit. We can also consider one-sided limits when we approach c from
only the right or the left. We write the left-hand limit as lim f ( x ) , and the right-hand
x!c "
limit as lim f ( x ) . Note that the value of the limit is independent of the value of the
x!c +
function at the input value c.
"1! x,
$
CONSIDER the function f (x) = #0,
$2 + x,
%
FIND lim f (x) and lim f (x) ?
x !0"
x <0
x = 0 . DRAW graph.
x >0
x !0+
sin x
sin x
and lim
.
x!0" x
x!0+ x
Two-sided limits
Note in the first example above the left-hand and right-hand limits both exist but
are different. Therefore, lim f (x) does not exist since 1 ! 2 .
FIND lim
x !0
For a function to have a limit at a point c the left-hand and right-hand limits must
sin x
sin x
sin x
exist and be equal to each other, e.g., lim"
= lim+
= 1 . Therefore, lim
= 1.
x!0
x!0
x!0
x
x
x
Continuity as it relates to the limit of a function.
We can now define a continuous function using limits. Assume that f ( x ) is
defined on an open interval containing x = c . Then f is continuous at x = c if
lim f ( x ) = f (c ) . If the limit does not exist, or if it exists but is not equal to f (c ) , we
x!c
say that f has a discontinuity (or is discontinuous) at x = c . E.g., g( x ) =
sin x
is not
x
sin x
= 1 , but g ( 0 ) ! 1 . On the other
x!0
x
hand, f ( x ) = 2x !1 is continuous at x = 3 , because lim ( 2x " 1) = f ( 3) = 5 .
continuous at x = 0 even though lim
x! 3
FIND a value of k such that the limit exists and find the limit.
x 2 + 5x + k
.
lim
x !"2
x+2
Find limit of a function algebraically, indeterminate form.
We say that f is indeterminate at x = c if we obtain an undefined expression for
0 !
x 2 " 2x " 3
is of
lim f ( x ) of the form , , ! " 0, ! # ! . E.g., the lim
0 !
x +1
x!"1
x!c
indeterminate form. Our strategy is to transform f algebraically, if possible, into a new
function that is continuous at x = c and then evaluate by substitution. Since
x 2 ! 2x ! 3 ( x ! 3)( x + 1)
x 2 " 2x " 3
= lim x " 3 since these two
=
= x ! 3 , then lim
x +1
x +1
x +1
x!"1
x!"1
functions agree everywhere except at x = !1 . Since lim x " 3 = "4 , the
x!"1
2
x " 2x " 3
= "4 , too.
x +1
x!"1
limit laws. The following basic limit laws make it easier to compute limits of
sums, multiples, products, and quotients of functions. Assume that lim f ( x ) and
lim
x!c
lim g( x ) exist. Then:
x!c
Sum Law: lim ( f ( x ) + g( x )) = lim f ( x ) + lim g( x )
x!c
x!c
x!c
Constant Multiple Law: For any number k, lim kf ( x ) = k lim f ( x )
x!c
"
%"
%
Product Law: lim f ( x ) g( x ) = $ lim f ( x )'$ lim g( x )'
# x!c
&# x!c
&
x!c
x!c
lim f ( x )
f ( x ) x!c
=
Quotient Law: If lim g( x ) " 0 , then lim
lim g( x )
x!c
x!c g( x )
x!c
1" f ( x )
.
x!1 g( x )
If lim f ( x ) = "2 and lim g( x ) = 3, FIND lim ( 2 f ( x ) " 3g ( x )) and lim
x!1
x!1
x!1